Lets say I have a message that is a sine wave at 1KHz. As I understand it, I can perform amplitude modulation by multiplying this signal with a carrier, lets say 10KHz, and this process is essentially DSB-SC modulation. Looking at the resulting signal in a spectrum analyzer I see a spectral line at 9KHz and 11Khz. My question is, why is there no spectral line at 10KHz? If you look at the waveform the 10KHz carrier wave is still clearly visible, it simply periodically inverts its phase 180 degrees. Is the fact that no spectral line appears in a frequency analyzer simply because it is being hidden, due to a limitation of how the spectrum analyzer works, effectively canceling out the 10KHz signal over time? If so, then is the carrier in DSB-SC truly suppressed? What does it actually mean for the carrier to be suppressed?

Edit. For clarification, I already know that the math says there should only be frequency content above and bellow the carrier frequency. Perhaps my confusion will become more clear if instead of modulating the carrier with a sine wave we modulate it with a square wave. Bellow is a picture of such a signal in Audacity. According to the spectrum analyzer this signal contains nothing at the original carrier frequency. This is despite the fact that the signal is literally made up of sections of the otherwise unmodified carrier, simply shifted 180 degrees periodically. It's not intuitive that a waveform that can be constructed from sections of a single sine wave of a particular frequency would not contain that frequency at all.

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It's a simple trigonometric identity:


Multiplying two sinusoids of different frequencies results in the sum of two sinusoids with the sum and the difference of the two frequencies.

So this type of AM (DSB-SC) really results in a suppressed carrier, i.e., the modulated signal has no component at the carrier frequency. This is true as long as the message signal has no DC component.

What happens is that the spectrum of the message signal is simply shifted to $\pm \omega_c$, so DC gets mapped to the carrier frequency $\omega_c$, and as long as there is no DC component in the message there is also no carrier component in the modulated signal.

EDIT: Your square wave example is perfect. A square wave is a piecewise constant function, so you could say it is just switched DC, or, paraphrasing your question, it's "made up of sections of otherwise unmodified DC". However, if the duration and amplitude of the positive and the negative part of the square wave are equal, then the corresponding signal's DC component is exactly zero. If you accept that, you must also accept that the signal in your example has no component at the carrier frequency.

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    $\begingroup$ I already know the trigonometric identity. It's probably the first thing you see on Wikipedia. That doesn't answer my question. It doesn't seem to intuitively follow that that should be the case. As I said in my question, if you look at the waveform the 10KHz carrier wave is still clearly visible, it simply periodically inverts its phase 180 degrees. So how do you explain that? If I took a tiny enough chunk of audio data, small enough that the phase does not have time to invert, then my intuition tells me that a 10KHz signal would be there. $\endgroup$ – Chris_F Dec 27 '19 at 13:46
  • $\begingroup$ Edit your question with the above comment, and the answers can address the gap in your intuition. It may seem counter-intuitive, but in the ideal case there really is no energy at the carrier frequency. Which means that your spectrum analyzer is working right. Yes, chopping the signal up into fine enough segments will regenerate the carrier - but to do that so that the carrier energy reappears means that your "chopping-up" signal would have to be synchronous with your modulating signal, which makes it a special case. $\endgroup$ – TimWescott Dec 27 '19 at 16:08
  • $\begingroup$ I updated my question. $\endgroup$ – Chris_F Dec 27 '19 at 16:57
  • $\begingroup$ @Chris_F: I've updated my answer to address your example of a square wave modulating a carrier. $\endgroup$ – Matt L. Dec 27 '19 at 18:56
  • $\begingroup$ A small window contains that frequency. But the sum of an infinite sequences of small windows of exact opposite polarity cancels out to zero. $\endgroup$ – hotpaw2 Dec 27 '19 at 22:44

If you window the two sidebands (so that they are not integer periodic in the window width), they will each generate their own (Sinc shaped) sidebands, which can occasionally constructively interfere to (re)create an artifact at the old center carrier frequency (as well as lots of other spectral splatter). The spectrum analyzer does not use such short windows (nor does a radio receiver), but your eye can, visually.

This will be true for the sum of any two pure sine waves at any two frequencies. A short enough window will make other frequencies seem to appear in a visual of just that window.


In AM DSB-SC modulation, carrier is suppressed. I think you ask "why is suppressed". The answer is power. In AM modulation you send signal with carrier, but on demodulation side you don't use carrier. The main purpose is reducing power dissipation.


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